The Envelope Theorem 1

John Nachbar

Washington University April 2, 2015

The Envelope Theorem

1 Introduction.

The Envelope theorem is a corollary of the Karush-Kuhn-Tucker theorem (KKT) that characterizes changes in the value of the objective function in response to changes in the parameters in the problem. For example, in a standard cost mini- mization problem for a firm, the Envelope theorem characterizes changes in cost in response to changes in input prices or output quantities.

2 The Envelope Theorem with no binding constraints.

Consider the following parameterized version of a differentiable MAX problem with no (binding) constraints,

max

f (x, q)

where q ∈ R

is a vector of parameters. The parametrized MIN problem is analo- gous. Let φ(q) give the (or at least, a) optimal x for a given q. The value function f

is defined by

f

(q) = f (φ(q), q).

The following is the Envelope theorem for this unconstrained problem. In theorem statement, D

f refers to the partial derivatives of f with respect to the x variables and D

f refers to the partial derivatives of f with respect to the q variables.

Theorem 1. Fix a parametrized differentiable MAX or MIN problem. Let U ⊆ R

be open, let φ : U → R

be differentiable, and suppose that, for every q ∈ U , φ(q) is a solution to the MAX or MIN problem. Let f

: U → R be defined by f

(q) = f (φ(q), q). Finally, suppose that the standard first order condition, D

f (φ(q), q) = 0, holds for every q ∈ U . Then for any q

∈ U , letting x

= φ(q

),

Df

(q

) = D

f (x

, q

).

1cbna. This work is licensed under the Creative Commons Attribution-NonCommercial- ShareAlike 4.0 License.

Proof. By the Chain Rule

Df

(q

) = D

f (x

, q

)Dφ(q

) + D

f (x

, q

).

Since D

f (x

, q

) = 0, the result follows. 

3 An example with no binding constraints.

Consider the cost minimization problem,

min

a + b s.t. √

ab ≥ y a, b ≥ 0

To interpret, a is capital, b is labor, and y is output. I assume (for simplicity) that the rental price of capital and the wage for labor both equal 1, so that total cost is a + b. If y > 0, any feasible a and b must be strictly positive, hence the non-negativity constraints do not bind at the solution. On the other hand, the production constraint √

ab ≥ y binds (since cost is strictly increasing in both a and b, and since the production function is continuous), hence I can rewrite this constraint as

b = y

a > 0.

Substituting, let the (modified) objective function be f (a, y) = a + y

a .

I can thus convert the above constrained minimization problem into an uncon- strained minimization problem

min

a +

.

(With the qualification that I have to check that indeed a > 0 at any proposed solution.) In terms of notation, here a is playing the role of “x” and y is playing the role of “q”.

2Formally, define r : U → R^{N +L}by r(q) = (φ(q), q). Then f^v(q) = f (r(q)). By the Chain Rule, Df^v(q^∗) = Df (x^∗, q^∗)Dr(q^∗)

=

Dxf (x^∗, q^∗) Dqf (x^∗, q^∗) 

 Dφ(q^∗) I



= Dxf (x^∗, q^∗)Dφ(q^∗) + Dqf (x^∗, q^∗).

as claimed. Many of the other Chain Rule applications are similar.

At y = y

, the first order condition D

f (a

, y

) = 0 gives, 1 − y

a

= 0,

hence y

= y

(which is strictly positive), hence φ(y) = a. For this particular application, call the value function C

(since the value function here is what is often called long-run cost). Then C

(y) = 2y.

Verifying the Envelope theorem,

DC

(y

) = D

f (a

, y

), since

D

f (a

, y

) = 2y

a

= 2.

Next, note that for a fixed (I won’t try to justify here in why a might be fixed), the value,

a + y

a ,

is the minimum cost of producing y (this is how I constructed the unconstrained problem in the first place). This can be interpreted as short-run cost, and accord- ingly I write,

C

(a, y) = a + y

a .

Short-run and long-run cost are equal when a = φ(y): C

(y) = C

(φ(y), y). By the Envelope theorem, then,

DC

(y

) = D

C

(a

, y

).

For any given value of a

, the graph of C

is strictly above that of C

(i.e., short- run cost is strictly higher than long-run cost) except at output level y

, since at that output level, the level of the fixed factor a is optimal. Moreover, when y = y

, so that φ(y

) = a

, the two graphs are tangent; their slopes are equal. One says that the graph of C

is the (lower) envelope of the graph of C

, hence the name Envelope theorem. Graphing C

and a few of the C

will help you visualize what is going on.

All of this generalizes. If C

is the short-run cost of producing a vector of outputs y given input prices and a fixed subvector of inputs a

, and if a

is optimal in the long-run when y = y

, then

DC

(y

) = D

C

(a

, y

).

4 The General Envelope Theorem.

Consider the following parameterized version of a differentiable MAX problem, max

f (x, q)

s.t. g

(x, q) ≤ 0, .. . g

(x, q) ≤ 0.

where q ∈ R

is a vector of parameters. For example, in a consumer maximization problem, the parameters are usually prices and income. The parameterized MIN problem is analogous. Again let φ(q) give the (or at least, a) optimal x for a given q and let the value function be defined by

f

(q) = f (φ(q), q).

Theorem 2 (Envelope Theorem). Fix a differentiable parameterized MAX or MIN problem. Let U ⊆ R

be open, let φ : U → R

be differentiable, and suppose that, for every q ∈ U , φ(q) is a solution to the MAX or MIN problem. Let f

: U → R be defined by f

(q) = f (φ(q), q). Let J (q) be the set of indices of constraints that are binding at φ(q). Finally, suppose that the conclusion of KKT holds for every q ∈ U . Fix any q

∈ U , let φ(q

) = x

, let J

= J (q

), and let λ

be the associated KKT multipliers. Then

Df

(q

) = D

f (x

, q

) − X

k∈J^∗

λ

D

g

(x

, q

).

Proof. By the Chain Rule (see the proof of Theorem 1), Df

(q

) = D

f (x

, q

)Dφ(q

) + D

f (x

, q

).

By KKT,

D

f (x

, q

) = X

k∈J

λ

D

g

(x

, q

).

Substituting,

Df

(q

) = X

k∈J

λ

D

g

(x

, q

)Dφ(q

) + D

f (x

, q

).

For any k ∈ J

, let γ

(q) = g

(φ(q), q). For a MAC problem, since k is binding at q

, γ

(q

) = 0 while for all other q ∈ U , γ

(q) ≤ 0 (since φ(q) must be feasible to be a solution). Therefore, q

maximizes γ

on U . (For a MIN problem, the argument is almost the same but q

minimizes γ

on U .) Therefore Dγ

(q

) = 0, hence, by the Chain Rule,

0 = Dγ

(q

) = D

g

(x

, q

)Dφ(q

) + D

g

(x

, q

).

Hence, D

g

(x

, q

)Dφ(q

) = −D

g

(x

, q

), hence, since λ

≥ 0, λ

D

g

(x

, q

)Dφ(q

) = −λ

D

g

(x

, q

).

Substituting this expression for λ

D

g

(x

, q

)Dφ(q

) into the above expression for Df

(q

) yields the result. 

5 An example with constraints.

Consider the problem

max

3

ln(x

) +

ln(x

) s.t. p · x ≤ m,

x ≥ 0

Where x ∈ R

, p ∈ R

and m ∈ R

. Think of this as a utility maximization problem with parameters being prices p  0 and income m > 0.

Grinding through the Kuhn-Tucker calculation, at prices p

and m

the solution is,

x

=

 m

/(3p

) (2m

)/(3p

)

 . Thus, at general (p, m), the solution function φ is given by,

φ(p, m) =

 m/(3p

) (2m)/(3p

)

 . Also

λ

= 1 m

and the other KKT multipliers (on the x ≥ 0 constraint) equal zero.

The value function is then determined by f

(p, m) = f (φ(p, m)), hence f

(p, m) = 1

3 ln  m 3p

 + 2

3 ln  2m 3p



= ln(m) − 1

3 ln(p

) − 2

3 ln(p

) + 1 3 ln  1

3

 + 2

3 ln  2 3

 .

Note that this makes some intuitive sense; in particular f

is increasing in m and decreasing in p

and p

.

By direct calculation, at the point (p

, m

),

Df

(p

, m

) =







∂f^v

∂p1

(p

, m

)

∂f^v

∂p2

(p

, m

)

∂f^v

∂m

(p

, m

)





 =





−1/(3p

)

−2/(3p

) 1/m



 .

On the other hand, by the Envelope theorem, since the parameters p and m don’t appear in the objective function, and with the first constraint written in standard form as g

(x, p, m) = p · x − m ≤ 0:

Df

(p

, m

) =





−λ

x

−λ

x

λ



 .

Substitute in the value of x

and λ

found above and you will confirm that these two expressions for Df

(p

, m

) are indeed equal.

6 The constraint term and the interpretation of the λ

.

Consider a MAX problem and focus on K parameters with the following special form. Parameter q

affects only constraint k and it does it additively:

g

(x) − q

≤ 0.

Increasing q

weakens the constraint. Then, from the Envelope theorem, at q

= 0, D

f

(0) = −λ

(−1) = λ

,

for any k ∈ J

. That is, λ

is the marginal value of relaxing binding constraint k.

The argument for MIN problems is almost identical.

For k / ∈ J , set λ

= 0. For some versions of KKT, this is actually a requirement

of KKT. For KKT as I have stated it, it is more in the nature of a convention. In

any event, the interpretation is that if constraint k is not binding then the marginal

value of relaxing constraint k is zero.